本帖最后由 tommywong 于 2020-6-27 13:28 编辑
我咁撚樣仲好
$\displaystyle S_m=\sum_{k=0}^\infty (-1)^k \binom{m-k}{k},~S_0=1,~S_1=1,~S_2=0$
$\displaystyle S_{m+1}=\sum_{k=0}^\infty (-1)^k \binom{m+1-k}{k}
=1+\sum_{k=1}^\infty (-1)^k \left(\binom{m-k}{k}+\binom{m-k}{k-1}\right)$
$\displaystyle =\sum_{k=0}^\infty (-1)^k \binom{m-k}{k}-\sum_{k=0}^\infty (-1)^k \binom{m-1-k}{k}$
$=S_m-S_{m-1}=S_{m-1}-S_{m-2}-S_{m-1}=-S_{m-2}$
$S_m=\begin{cases}
1 & m\equiv 0\pmod{6}\\
1 & m\equiv 1\pmod{6}\\
0 & m\equiv 2\pmod{6}\\
-1 & m\equiv 3\pmod{6}\\
-1 & m\equiv 4\pmod{6}\\
0 & m\equiv 5\pmod{6}
\end{cases}
=\begin{cases}
(-1)^m & m\equiv 0\pmod{3}\\
-(-1)^m & m\equiv 1\pmod{3}\\
0 & m\equiv 2\pmod{3}\\
\end{cases}$
$\displaystyle \frac{1}{m-k}\binom{m-k}{k}=\frac{1}{m}\left(\binom{m-k}{k}+\binom{m-2-(k-1)}{k-1}\right)$
$\displaystyle \sum_{k=0}^\infty (-1)^k \frac{1}{m-k}\binom{m-k}{k}
=\frac{1}{m}\sum_{k=0}^\infty (-1)^k\binom{m-k}{k}
+\frac{1}{m}\sum_{k=1}^\infty (-1)^k\binom{m-2-(k-1)}{k-1}$
$\displaystyle =\frac{S_m-S_{m-2}}{m}=\frac{S_{m+1}+S_m}{m}
=\begin{cases}
\frac{2}{m} & m\equiv 0\pmod{6}\\
\frac{1}{m} & m\equiv 1\pmod{6}\\
\frac{-1}{m} & m\equiv 2\pmod{6}\\
\frac{-2}{m} & m\equiv 3\pmod{6}\\
\frac{-1}{m} & m\equiv 4\pmod{6}\\
\frac{1}{m} & m\equiv 5\pmod{6}\\
\end{cases}
=\begin{cases}
\frac{2(-1)^m}{m} & m\equiv 0\pmod{3}\\
\frac{-(-1)^m}{m} & m\equiv 1\pmod{3}\\
\frac{-(-1)^m}{m} & m\equiv 2\pmod{3}\\
\end{cases}$ |